Question: Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(tr^{-1})^{-3}}}{{(t^{-2}r^{-2})^{-4}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(tr^{-1})^{-3} = (t)^{-3}(r^{-1})^{-3}}$ On the left, we have ${t}$ to the exponent ${-3}$ . Now ${1 \times -3 = -3}$ , so ${(t)^{-3} = t^{-3}}$ Apply the ideas above to simplify the equation. $\dfrac{{(tr^{-1})^{-3}}}{{(t^{-2}r^{-2})^{-4}}} = \dfrac{{t^{-3}r^{3}}}{{t^{8}r^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{-3}r^{3}}}{{t^{8}r^{8}}} = \dfrac{{t^{-3}}}{{t^{8}}} \cdot \dfrac{{r^{3}}}{{r^{8}}} = t^{{-3} - {8}} \cdot r^{{3} - {8}} = t^{-11}r^{-5}$